If -3-2-k-k- then the unit vector parallel to the vector - is

Let $\to a =^i +^j +^k, \to b =^i -^j +^k$ and $\to c =^i -^j -^k$ be three vectors. A vector $\to v$ in the plane of
$\to a$ and $\to b$ , whose projection on $\to c$ is $\frac{1}{\sqrt{3}}$ , is given by

Q. Let

\to α = 3^i +^j

and

\to β = 2^i -^j + 3^k

. If

\to β =_{1} - β_{2}

, where

_{1}

is parallel to

\to α

and

_{2}

is perpendicular to

\to α

, then

_{1} \times_{2}

is equal to?

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If →α=3^i−2^i+^k,→β=−^i+^i+^k, then the unit vector parallel to the vector (→α+→β) is

The correct option is A (23)^i−(13)^i+(23)^kWe get α+β=2^i−^j+2^k|α+β|=3So Unit vector is α+β|α+β|=13(2^i−^j+2^k)

If $\to α = 3^i - 2^i +^k, \to β = -^i +^i +^k$ , then the unit vector parallel to the vector $(\to α + \to β)$ is

The correct option is A $(\frac{2}{3})^i - (\frac{1}{3})^i + (\frac{2}{3})^k$
We get $α + β = 2^i -^j + 2^k$
$| α + β | = 3$
So Unit vector is $\frac{α + β}{| α + β |} = \frac{1}{3} (2^i -^j + 2^k)$